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Viktor [21]
3 years ago
8

An object begins with a speed of 20 meters per second and slows down to a speed of 10 meters per second over a time of 5 seconds

. Calculate the average speed of the object during these 5 seconds.
can someone tell me how to do this please?
Physics
1 answer:
Andreyy893 years ago
3 0

Answer:

V = 6 m/s

Explanation:

Given that,

Initial speed of an object is 20 m/s

Final speed of an object is 10 m/s

Time, t = 5 s

We need to find the average speed of the object during these 5 seconds. Let it is equal to V. Here, time is same. The average speed is given by :

V=\dfrac{x+y}{2}\\\\\text{Putting values}\\\\V=\dfrac{20+10}{5}\\\\V=6\ m/s

So, the average speed of the object is 6 m/s.

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The earth has radius R. A satellite of mass 100 kg is in orbit at an altitude of 3R above the earth's surface. What is the satel
Vera_Pavlovna [14]

Answer:

W= 61.3 N

Explanation:

The only force acting on the satellite is the one due to the attraction from Earth, which obeys the Newton's Universal Law of Gravitation, as follows:

Fg =G*ms*me / (res)²

This force, also obeys the Newton's 2nd Law, so we can write the following equation:

G*ms*me*/ (res)² = ms* a = ms*g

We call to the product of the mass times the acceleration caused by gravity (g), the weight of this mass, so we can write as follows:

G*ms*me / (res)² = ms*g = W (1)

where G = 6.67*10⁻11 N*m²/kg², ms= 100 kg, me= 5.97*10²⁴ kg, and

res=  4 *re = 4*6.37*10⁶ m.

Replacing all these known values in (1), we get the value of W:

W =(( 6.67*5.97/(4*6.37)²) *( 10⁻¹¹ * 10²⁴ /10¹²) )* 100 N = 61.3 N

6 0
3 years ago
A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th
irina [24]

The angular momentum is defined as,

L=I\omega

Acording to this text we know for conservation of angular momentum that

L_i=L_f

Where L_iis initial momentum

L_f is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

L_i=mvl

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

L_f=(I_b+I_s)\omega

Substituting

L_f=(ml^2+\frac{10ml^2}{3})\omega

L_f=\frac{13}{10}ml^2w

m(0.65)l=\frac{13}{10}ml^2 \omega

\omega=\frac{1}{2l}

Applying conservative energy equation we have

\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'

\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)

Replacing the values and solving

l=\frac{13}{0.54g}

Substituting

l=\frac{13}{0.54(9.8)}

l=2.45cm

7 0
3 years ago
Describe how a battery works
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Magic, Nah im just kidding. A battery has two parts, the anode and the cathode. Which anode is positive and cathode is negative, which they are connected to the electrolyte. Once they are connected to a device they once start working from separate ends. Which is the flow of energy 
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3 years ago
A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, what is the mass of
Deffense [45]

Answer:

the mass of the air in the classroom = 2322 kg

Explanation:

given:

A classroom is about 3 meters high, 20 meters wide and 30 meters long.

If the density of air is 1.29 kg/m3

find:

what is the mass of the air in the classroom?

density = mass / volume

where mass (m) = 1.29 kg/m³

volume = 3m x 20m x 30m = 1800 m³

plugin values into the formula

  1.29 kg/m³   =  <u>      mass    </u>

                             1800 m³

mass =  1.29 kg/m³  ( 1800 m³ )

mass = 2322 kg

therefore,

the mass of the air in the classroom = 2322 kg

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3 years ago
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Suppose an astronaut has landed on Planet * Fully equipped, the astronaut has a
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From the calculations, the value of the acceleration due to gravity is 0.38 m/s^2.

<h3>What is weight?</h3>

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Thus;

When;

mass = 120 kg

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acceleration due to gravity = 46 N/120 kg

=0.38 m/s^2

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